How would you calculate the pH of a substance after a titration of 10.19mL 0.10M HCl with 10.99mL 0.093M NaOH?

1 Answer
Dec 22, 2015

Answer:

#"pH" = 10.15#

Explanation:

You're mixing hydrochloric acid, #"HCl"#, a strong acid, and sodium hydroxide, #"NaOH"#, a strong base, which means that the pH of the resulting solution will depend on whether or not you're dealing with a complete neutralization.

The balanced chemical equation for this neutralization reaction looks like this

#"HCl"_text((aq]) + "NaOH"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O"_text((l])#

In order to have a complete neutralization, you need to add equal numbers of moles of strong acid and weak base.

If you have more acid than base, the resulting solution will be acidic, since it will contain an excess of hydronium ions, #"H"_3"O"^(+)#.

Likewise, if you have more base than acid, the resulting solution will be basic, since it will contain an excess of hydroxide anions, #"OH"^(-)#.

Use the molarities and volumes of the two solutions to determine how many moles of each you;'re adding together

#color(blue)(c = n/V implies n = c * V)#

#n_(HCl) = "0.10 M" * 10.19 * 10^(-3)"L" = "0.001019 moles HCl"#

#n_(NaOH) = "0.093 M" * 10.99 * 10^(-3)"L" = "0.001022 moles NaOH"#

Notice that you have more moles of sodium hydroxide than you do of hydrochloric acid. This tells you that the reaction will completely consume the acid and leave you with

#n_(NaOH) = 0.001022 - 0.001019 = "0.0000030 moles NaOH"#

Sodium hydroxide dissociates completely in aqueous solution, so you can say that

#n_(OH^(-)) = n_(NaOH) = "0.0000030 moles OH"^(-)#

The total volume of the resulting solution will be

#V_"total" = "10.19 mL" + "10.99 mL" = "21.18 mL"#

The molarity of the hydroxide anions will be

#c = "0.0000030 moles"/(21.18 * 10^(-3)"L") = 1.42 * 10^(-4)"M"#

As you know, a solution's pOH is equal to

#"pOH" = - log( ["OH"^(-)])#

In this case.

#"pOH" = - log(1.42 * 10^(-4)) = 3.85#

The pH of the solution will thus be

#"pH" + "pOH" = 14#

#"pH" = 14 - 3.85 = color(green)(10.15)#